Shunt Resistance for Ammeter Conversion

Low resistance S in parallel with galvanometer of resistance G allows current (I − Ig) to bypass, giving full-scale deflection at current I. Converted ammeter has very low net resistance.

Problem

A galvanometer with resistance $G$ gives full-scale deflection at current $I_g$. We want full-scale deflection when the total circuit current is $I \gg I_g$.

Setup

Connect a low resistance $S$ (shunt) in parallel with the galvanometer. Current $I_g$ passes through the galvanometer; $(I - I_g)$ bypasses through the shunt.

Derivation

At full-scale deflection, voltage across both is equal:

$$I_g G = (I - I_g)S$$ $$\boxed{S = \frac{G\,I_g}{I - I_g}}$$

Net Resistance of Ammeter

$$R_A = \frac{GS}{G + S} = \frac{G \cdot \frac{GI_g}{I-I_g}}{G + \frac{GI_g}{I-I_g}} = \frac{GI_g}{I}$$

Since $I \gg I_g$, $R_A \ll G$ — ammeter has very low resistance, causing minimal voltage drop when inserted in series in the circuit.